What is the solution of the system of inequalities y>x^2+6x+10 y<-x^2-8x-14

if a rug was 5 times as long as thia record sheet, how long would it be? measure the length of this record sheet in centimeters.

vlada-n [284]

<u>Answer: </u>

If a rug was 5 times as long as this record sheet. The length of this record sheet in centimeters is 100x centimeter

<u>Solution: </u>

Given that a rug was 5 times than the record sheet.

Let assume the length of the record sheet = x units

Then from given information,

Length of the rug = 5x units

Conversion of measure of record sheet in centimeter completely depends upon in which unit the length is given.

Suppose the given length is in meters, than need to multiply the given length by 100 to get the length in centimeter that is

x meter = $x \times 100$ = 100x cm

Conversion factor will change accordingly to the unit of given length which needs to be convert into centimeter.

4 0

3 years ago

Solve 6x + c = k for x

Thepotemich [5.8K]

Answer:

x= k/6 - c/6

Step-by-step explanation:

5 0

3 years ago

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What is the slope of a line perpendicular to the line whose equation is

seropon [69]

Answer:

1

Step-by-step explanation:

x+y=-9

y=-x-9

So slope of original line is -1.

The line perpendicular to the original line is -1/-1 = 1

5 0

3 years ago

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Fifty percent of students in your school own a bike. Of a sample of students in your school, the ratio of students who own a bik

Svetradugi [14.3K]

Answer:

Yes

Step-by-step explanation:

If 4 is half of 8, and 50% is half of 100%, then 4:8 is an accurate ratio depecting that half the school owns a bike.

8 0

3 years ago

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What is the antiderivative of 3x/((x-1)^2)

Maslowich

Answer:

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Step-by-step explanation:

Given

$\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=3\cdot \int \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Apply\:u-substitution:}\:u=x-1$

$=3\cdot \int \frac{u+1}{u^2}du$

$\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}$

$=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)$

as

$\int \frac{1}{u}du=\ln \left|u\right|$ ∵ $\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\int \frac{1}{u^2}du=-\frac{1}{u}$ ∵ $\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

so

$=3\left(\ln \left|u\right|-\frac{1}{u}\right)$

$\mathrm{Substitute\:back}\:u=x-1$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Therefore,

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

4 0

3 years ago